December 9, 2018 | Author: Dex Valdez | Category: Number Theory, Equations, Abstract Algebra, Mathematical Problem Solving, Mathematical Analysis

#### Description

Metrobank-MTAP-DepEd Math Challenge 2013 Individual Finals •  Second Year •  Category A •  Answer Key Part I 1. 3x + 2y 2 y + 6 = 0

4. x/( x/(x + 1)

7. 20

10. 6

13.   120 km

2. 70

5. 5

8.   10 kph

11. (5 − y )/4

14. 4

3. 7/198

6. 50

9. 22

12. 4/(x2 + 5x 5x + 6) 15. 4

Part II 1. For a ≥  1, we must have 1  ≤  b ≤  6.   (1 pt) We pt)  We check for every b. If  b  = 1, there are 10 a’s that satisfy the inequality inequality. If b If  b  = 2, there are 9 a’s. If b If  b  = 3, there are 7 a’s. If  b  b  = 4, there are 5 a’s. If b If  b  = 5, there are 4 a’s. Finally Finally, if  b  = 6, there are 2 a’s. All in all, there are 37 ordered pairs.  (2 pts) for complete veriﬁcation, and (1 pt) for at least three veriﬁcations 2. Let U  = {1, 2, 3, . . . , 999}, A  be the subset of  U  consisting U  consisting of all multiples of 6, and B   the subset of  U  consisting of multiples of 7. Using Venn Diagram, we have

|A ∪ B |  =  |A| + |B | − |A ∩ B |  = 166 + 142 − 23 = 285. 285. (2 pts) pts) Thus, there are 999 − 285 = 714 positive integers less than 1000 that are not divisible by 6 or 7.  (1 pt) 3. Let m Let  m be  be the number number of days days Marvin can ﬁnish the work alone. alone. Then Vincent Vincent can ﬁnish the work in m in m + 2 days.   (1 pt) We pt)  We solve for m  in the equation 1 1 1 + = 12 (1 pt) m m + 2 5

=⇒

m = 4.  (1 pt)

4. Let , w, and d  be the length, length, width, and diagonal of the rectangle, rectangle, respectively respectively.. Then  = 2w  + 4 and 2 2  =   +  + 2 = 2w 2w + 6.  (1 pt) By pt)  By Pythagorean Theorem, we get w get  w +  = d 2 , or d  =  w2 + (2w (2w + 4) 2 = (2w (2w + 6) 2

=⇒

w  = 10 10..   (1 pt)

Thus,  Thus,    = 240, and the area of the rectan rectangle gle is 240 cm2 .  (1 pt) 5. Let r  be one of the roots of the given quadratic equation. Then the other root is 2r 2r  + 3.   (1 pt) By pt)  By the relationship relationshipss between between the roots and the coeﬃcients, coeﬃcients, we have r have r +  + (2r (2r + 3) = − p and  p  and  r(2  r (2rr + 3) = q .  (1 pt) 1 The ﬁrst equation gives 3r 3r  =  − p − 3 or r or  r  = 3 (− p − 3). Substituting this to the second equation, we get 2

2

q  =  = 2r + 3r 3r  = 2

 − p − 3  3

+ (− p − 3) =

2 p2 + 3 p 3 p − 9 .   (1 pt) 9

Part III 1. Let 1 and 2   be the lengths lengths of the sides sides of the two two square squares. s. Then Then we solve solve the system system of equatio equations: ns: 2 2 41 + 42  = 44 and  and   1 + 2  = 65.  ((2 pts) for creating the system) This system) This system gives  gives   1  = 4 and  and   2  = 7 (or  (or   1  = 7 and  and   2  = 4).   ((3 pts) for completely solving the system) 2. Let  Let   and  and w  w be  be the length and width of the rectangle, respectively. Then we solve the system of equations: ( + 2)(w 2)(w  − 1) = w and w  and ( ( − 2)(w 2)(w + 4) = w. w.   ((2 pts) for creating the system) This system)  This system gives   = 10 10/ /3 cm and w and  w  = 8/3 cm,  cm,   ((2 pts) for completely solving the system) which system)  which give 80/ 80/9 cm2 for the area.  ((1 pt) for computing the area) 3. Let a Let  a  and  b  be the bids of Company A Company  A  and Company B Company  B , respectively. Then we have the following system of equations: a + b  +  b  = 90000 and 0. 0.9b −  0.  0 .95 95a a = (b −  a)  a)  −  3000.   ((2 pts) for creating the system) Solving the system yields a  = 40 40,, 000 and b  = 50 50,, 000 000   ((3 pts) for completely solving the system); system); that is, A  bid Php 40, 40, 000, while B while  B  bid Php 50, 50, 000.